Is it true that stabilizer in $O(n)$ of a rank $k$ matrix is isomorphic to $O(n-k)$?

Question

Let $X\in M_{n, k}(\mathbb R)$ such that $\textrm{rank}(X)=k$ and,$$O(n)_X:=\{A\in O(n): AX=X\}.$$ Notice $O(n)_X$ is a subgroup of $O(n)$. Is it true that $O(n-k)\cong O(n)_X$? Here $O(n)=\{A\in M_n(\mathbb R): A^TA=I\}$..

Answer 1

The answer is yes (except when $n=k$, where $O(0)$ is undefined). Geometrically, this is because $A$ fixes all $k$-columns of $X$ and it is free to perform any orthogonal transform on the orthogonal complement of the column space of $X$.